Mechanical Properties of Fluids

Physics Part II

Physics

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Water rises to a height 4.8cm in a certain capillary tube. When the same tube is dipped in some another liquid it falls by 12.5cm. Compare the surface tension of water and liquid. Density of liquid is 9.6 gm/cc and angle of contact of liquid is 120°.

Given,

Rise in water = 4.8 cm

Fall in water when tube is dipped in some other liquid = 12.5 cm

Density of liquid = 9.6 gm/cc

Angle of contact of liquid = 120

The height to which the liquid rises or falls in the capillary tube is given by the relation,

$\mathrm{h}=\frac{2\mathrm{T}\mathrm{cos\theta}}{\mathrm{rpg}}$

The above relation can be simplified to

$\mathrm{T}=\frac{\mathrm{rgph}}{2\mathrm{cos}\mathrm{\theta}}....\left(1\right)$

Let T

By using the equation (1), we have

$\frac{{\mathrm{T}}_{1}}{{\mathrm{T}}_{2}}=\frac{{\mathrm{\rho}}_{1}{\mathrm{h}}_{1}}{{\mathrm{\rho}}_{2}{\mathrm{h}}_{2}}\frac{{\mathrm{cos\theta}}_{2}}{{\mathrm{cos\theta}}_{1}}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Here},{\mathrm{\rho}}_{1}=1\mathrm{gm}/\mathrm{cc},{\mathrm{h}}_{1}=4.8\mathrm{cm},{\mathrm{\theta}}_{1}=0\xb0\phantom{\rule{0ex}{0ex}}{\mathrm{\rho}}_{2}=9.6\mathrm{gm}/\mathrm{cc},{\mathrm{h}}_{2}=-12.5\mathrm{cm},{\mathrm{\theta}}_{2}=120\xb0\phantom{\rule{0ex}{0ex}}\therefore \frac{{\mathrm{T}}_{1}}{{\mathrm{T}}_{2}}=\frac{1\times 4.8}{9.6\times (-1.25)}\frac{\mathrm{cos}120\xb0}{\mathrm{cos}0\xb0}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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Do intermolecular or inter-atomic forces follow inverse square law?

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Why solids have definite shape while liquids do not have definite shape?

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